Question

1684. 统计一致字符串的数目

English Version

题目描述

给你一个由不同字符组成的字符串 allowed 和一个字符串数组 words 。如果一个字符串的每一个字符都在 allowed 中，就称这个字符串是 一致字符串 。

请你返回 words 数组中 一致字符串 的数目。

 

示例 1：

输入：allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
输出：2
解释：字符串 "aaab" 和 "baa" 都是一致字符串，因为它们只包含字符 'a' 和 'b' 。

示例 2：

输入：allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
输出：7
解释：所有字符串都是一致的。

示例 3：

输入：allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
输出：4
解释：字符串 "cc"，"acd"，"ac" 和 "d" 是一致字符串。

 

提示：

• 1 <= words.length <= 104
• 1 <= allowed.length <= 26
• 1 <= words[i].length <= 10
• allowed 中的字符 互不相同 。
• words[i] 和 allowed 只包含小写英文字母。

解法

方法一：哈希表或数组

一种比较直接的思路是，用哈希表或数组 $s$ 记录 allowed 中的字符。然后遍历 words 数组，对于每个字符串 $w$，判断其是否由 allowed 中的字符组成。若是，答案加一。

时间复杂度 $O(m)$，空间复杂度 $O(C)$。其中 $m$ 为所有字符串的总长度，而 $C$ 为 allowed 字符集的大小。本题中 $C \leq 26$。

class Solution:
  def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
    s = set(allowed)
    return sum(all(c in s for c in w) for w in words)

class Solution {
  public int countConsistentStrings(String allowed, String[] words) {
    boolean[] s = new boolean[26];
    for (char c : allowed.toCharArray()) {
      s[c - 'a'] = true;
    }
    int ans = 0;
    for (String w : words) {
      if (check(w, s)) {
        ++ans;
      }
    }
    return ans;
  }

  private boolean check(String w, boolean[] s) {
    for (int i = 0; i < w.length(); ++i) {
      if (!s[w.charAt(i) - 'a']) {
        return false;
      }
    }
    return true;
  }
}

class Solution {
public:
  int countConsistentStrings(string allowed, vector<string>& words) {
    bitset<26> s;
    for (auto& c : allowed) s[c - 'a'] = 1;
    int ans = 0;
    auto check = [&](string& w) {
      for (auto& c : w)
        if (!s[c - 'a']) return false;
      return true;
    };
    for (auto& w : words) ans += check(w);
    return ans;
  }
};

func countConsistentStrings(allowed string, words []string) (ans int) {
  s := [26]bool{}
  for _, c := range allowed {
    s[c-'a'] = true
  }
  check := func(w string) bool {
    for _, c := range w {
      if !s[c-'a'] {
        return false
      }
    }
    return true
  }
  for _, w := range words {
    if check(w) {
      ans++
    }
  }
  return ans
}

function countConsistentStrings(allowed: string, words: string[]): number {
  const set = new Set([...allowed]);
  const n = words.length;
  let ans = n;
  for (const word of words) {
    for (const c of word) {
      if (!set.has(c)) {
        ans--;
        break;
      }
    }
  }
  return ans;
}

impl Solution {
  pub fn count_consistent_strings(allowed: String, words: Vec<String>) -> i32 {
    let n = words.len();
    let mut make = [false; 26];
    for c in allowed.as_bytes() {
      make[(c - b'a') as usize] = true;
    }
    let mut ans = n as i32;
    for word in words.iter() {
      for c in word.as_bytes().iter() {
        if !make[(c - b'a') as usize] {
          ans -= 1;
          break;
        }
      }
    }
    ans
  }
}

int countConsistentStrings(char* allowed, char** words, int wordsSize) {
  int n = strlen(allowed);
  int make[26] = {0};
  for (int i = 0; i < n; i++) {
    make[allowed[i] - 'a'] = 1;
  }
  int ans = wordsSize;
  for (int i = 0; i < wordsSize; i++) {
    char* word = words[i];
    for (int j = 0; j < strlen(word); j++) {
      if (!make[word[j] - 'a']) {
        ans--;
        break;
      }
    }
  }
  return ans;
}

方法二：位运算

我们也可以仅用一个整数来表示每个字符串中字符的出现情况。其中，整数的二进制表示中的每一位表示一个字符是否出现。

我们简单地定义一个函数 $f(w)$，这个函数可以将一个字符串 $w$ 转换为一个整数。整数的二进制表示中的每一位表示一个字符是否出现。例如，字符串 ab 可以转换为整数 $3$，即二进制表示为 $11$。字符串 abd 可以转换为整数 $11$，即二进制表示为 $1011$。

回到题目上，判断一个字符串 $w$ 是否由 allowed 中的字符组成，就可以转换为：判断 $f(allowed)$ 和 $f(w)$ 进行按位或运算后的结果是否等于 $f(allowed)$。若是，答案加一。

时间复杂度 $O(m)$，其中 $m$ 为所有字符串的总长度。空间复杂度 $O(1)$。

class Solution:
  def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
    def f(w):
      return reduce(or_, (1 << (ord(c) - ord('a')) for c in w))

    mask = f(allowed)
    return sum((mask | f(w)) == mask for w in words)

class Solution {
  public int countConsistentStrings(String allowed, String[] words) {
    int mask = f(allowed);
    int ans = 0;
    for (String w : words) {
      if ((mask | f(w)) == mask) {
        ++ans;
      }
    }
    return ans;
  }

  private int f(String w) {
    int mask = 0;
    for (int i = 0; i < w.length(); ++i) {
      mask |= 1 << (w.charAt(i) - 'a');
    }
    return mask;
  }
}

class Solution {
public:
  int countConsistentStrings(string allowed, vector<string>& words) {
    auto f = [](string& w) {
      int mask = 0;
      for (auto& c : w) mask |= 1 << (c - 'a');
      return mask;
    };
    int mask = f(allowed);
    int ans = 0;
    for (auto& w : words) ans += (mask | f(w)) == mask;
    return ans;
  }
};

func countConsistentStrings(allowed string, words []string) (ans int) {
  f := func(w string) (mask int) {
    for _, c := range w {
      mask |= 1 << (c - 'a')
    }
    return
  }

  mask := f(allowed)
  for _, w := range words {
    if (mask | f(w)) == mask {
      ans++
    }
  }
  return
}

function countConsistentStrings(allowed: string, words: string[]): number {
  const helper = (s: string) => {
    let res = 0;
    for (const c of s) {
      res |= 1 << (c.charCodeAt(0) - 'a'.charCodeAt(0));
    }
    return res;
  };
  const mask = helper(allowed);
  let ans = 0;
  for (const word of words) {
    if ((mask | helper(word)) === mask) {
      ans++;
    }
  }
  return ans;
}

impl Solution {
  fn helper(s: &String) -> i32 {
    let mut res = 0;
    for c in s.as_bytes().iter() {
      res |= 1 << ((c - b'a') as i32);
    }
    res
  }

  pub fn count_consistent_strings(allowed: String, words: Vec<String>) -> i32 {
    let mask = Self::helper(&allowed);
    let mut ans = 0;
    for word in words.iter() {
      if (mask | Self::helper(word)) == mask {
        ans += 1;
      }
    }
    ans
  }
}

int helper(char* s) {
  int res = 0;
  int n = strlen(s);
  for (int i = 0; i < n; i++) {
    res |= 1 << (s[i] - 'a');
  }
  return res;
}

int countConsistentStrings(char* allowed, char** words, int wordsSize) {
  int mask = helper(allowed);
  int ans = 0;
  for (int i = 0; i < wordsSize; i++) {
    if ((mask | helper(words[i])) == mask) {
      ans++;
    }
  }
  return ans;
}